Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{x^2 - 4x - 5}{-2x + 10} \div \dfrac{7x + 7}{x + 6} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{x^2 - 4x - 5}{-2x + 10} \times \dfrac{x + 6}{7x + 7} $ First factor the quadratic. $q = \dfrac{(x + 1)(x - 5)}{-2x + 10} \times \dfrac{x + 6}{7x + 7} $ Then factor out any other terms. $q = \dfrac{(x + 1)(x - 5)}{-2(x - 5)} \times \dfrac{x + 6}{7(x + 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (x + 1)(x - 5) \times (x + 6) } { -2(x - 5) \times 7(x + 1) } $ $q = \dfrac{ (x + 1)(x - 5)(x + 6)}{ -14(x - 5)(x + 1)} $ Notice that $(x - 5)$ and $(x + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(x + 1)}(x - 5)(x + 6)}{ -14(x - 5)\cancel{(x + 1)}} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $q = \dfrac{ \cancel{(x + 1)}\cancel{(x - 5)}(x + 6)}{ -14\cancel{(x - 5)}\cancel{(x + 1)}} $ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ $q = \dfrac{x + 6}{-14} $ $q = \dfrac{-(x + 6)}{14} ; \space x \neq -1 ; \space x \neq 5 $